Question Number 172730 by cortano1 last updated on 30/Jun/22
Answered by mr W last updated on 30/Jun/22
Commented by mr W last updated on 30/Jun/22
$$\frac{\mathrm{sin}\:\left({x}+\mathrm{45}\right)}{{AD}}=\frac{\mathrm{sin}\:{x}}{{BD}} \\ $$$$\frac{{AD}}{\mathrm{sin}\:\left(\mathrm{45}−\mathrm{15}\right)}=\frac{{DC}}{\mathrm{sin}\:\mathrm{15}} \\ $$$$\frac{\mathrm{sin}\:\left({x}+\mathrm{45}\right)}{\mathrm{sin}\:\left(\mathrm{45}−\mathrm{15}\right)}=\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:\mathrm{15}} \\ $$$$\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:{x}}}{\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{15}}=\mathrm{1} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:{x}}=\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{15}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}\right)}{\mathrm{4}}=\mathrm{1}+\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\mathrm{30}°\:\checkmark \\ $$
Commented by Tawa11 last updated on 01/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$